Question: Simplify and expand the following expression: $ \dfrac{x - 3}{4x - 6}+\dfrac{1}{x + 2} $
In order to add expressions, they must have a common denominator. Get both fractions over a common denominator of $(4x - 6)(x + 2)$ Multiply the first term by $\dfrac{x + 2}{x + 2}$ $ \begin{align*} \dfrac{x - 3}{4x - 6} \times \dfrac{x + 2}{x + 2} & = \dfrac{(x - 3)(x + 2)}{(4x - 6)(x + 2)} \\ & = \dfrac{x^2 - x - 6}{(4x - 6)(x + 2)}\end{align*} $ Multiply the second term by $\dfrac{4x - 6}{4x - 6}$ $ \begin{align*} \dfrac{1}{x + 2} \times \dfrac{4x - 6}{4x - 6} & = \dfrac{(1)(4x - 6)}{(x + 2)(4x - 6)} \\ & = \dfrac{4x - 6}{(x + 2)(4x - 6)}\end{align*} $ Now we have: $ = \dfrac{x^2 - x - 6}{(4x - 6)(x + 2)} + \dfrac{4x - 6}{(x + 2)(4x - 6)} $ Now both terms have a common denominator we can simply add the numerators: $ = \dfrac{x^2 - x - 6 + 4x - 6}{(4x - 6)(x + 2)} $ $ = \dfrac{x^2 + 3x - 12}{(4x - 6)(x + 2)}$ Expand the denominator: $ = \dfrac{x^2 + 3x - 12}{4x^2 + 2x - 12}$